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Viktor [21]
3 years ago
8

An object begins with a speed of 20 meters per second and slows down to a speed of 10 meters per second over a time of 5 seconds

. Calculate the average speed of the object during these 5 seconds.
can someone tell me how to do this please?
Physics
1 answer:
Andreyy893 years ago
3 0

Answer:

V = 6 m/s

Explanation:

Given that,

Initial speed of an object is 20 m/s

Final speed of an object is 10 m/s

Time, t = 5 s

We need to find the average speed of the object during these 5 seconds. Let it is equal to V. Here, time is same. The average speed is given by :

V=\dfrac{x+y}{2}\\\\\text{Putting values}\\\\V=\dfrac{20+10}{5}\\\\V=6\ m/s

So, the average speed of the object is 6 m/s.

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3 0
3 years ago
A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho
fiasKO [112]

Answer:

v_f=8.17\frac{m}{s}

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}

The box is initially at rest, so v_i=0. Solving for v_f:

v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}

5 0
2 years ago
How many number of musicians constitute a big band?
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6 0
3 years ago
How would the plane strain fracture toughness of a metal be expected to change with rising temperature?
aksik [14]

Answer:

Increase

Explanation:

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4 0
3 years ago
Below is a circuit schematic of sources and resistors (Figure 3). VS = 10V , R1 = 100Ω, R2 = 50Ω, R3 = 25Ω, IS = 2A. Calculate t
lorasvet [3.4K]

Answer:

V_3\approx 4.28\,\,V

I_1=0.0572\,\,amps

I_3\approx 0.171\,\,amps

Explanation:

Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.

So we first find the equivalent resistance for the two resistors in parallel:

\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega

By knowing this, we can estimate the total current through the circuit,:

Vs=I\,*\,(\frac{100}{3} +25)\\10=I\,*\,\frac{175}{3} \\I=\frac{30}{175} \,amps

So approximately 0.17  amps

and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

V_3=\frac{30}{175} *\,25=\frac{30}{7} \approx 4.28\,\,V

So now we know that the potential drop across the parellel resistors must be:

10 V -  4.28 V = 5.72 V

and with this info, we can calculate the current through R1 using Ohm's Law:

I_1=\frac{V_1}{R_1} =\frac{5.72}{100} =0.0572\,\,amps

4 0
3 years ago
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